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Talk:Notation Array Notation
I hate to be the bearer of bad news, but this notation just doesn't work. You can't just stitch together a few notations, label them levels 1 through 5 and expect the 6th and beyond members to be defined. in other words, what i'm saying is, + */^ -> A don't form a sequence S such that you can simply invoke Sn for n>5 In fact, you are comitting one of the sins Sbiis warned us about in his Introduction to Googology which i will allow myself to quote for you "2.' Don't assume that some vague sequence HAS to have a defined continuation'. Sometimes people make this mistake when they are trying to "break out" of the system. " Chronolegends (talk) 06:24, October 21, 2016 (UTC) :Right, 6th level or above isnt defined. You cant form the sequence above. AarexWikia04 - 11:52, October 21, 2016 (UTC) I realized that I forgot to add Greek Notation to the main wiki page, since my notation depends on it to be definable. Otherwise, it is an arbitrary function. Here are the rules anyways which I will copy to a new page on Greek Notation later. Please check both Greek Notation page and NaN to realize the full definition: Each level of the first bracket number is an operational recursion of the previous function. For example, if you want to make large towers of exponents, you wouldn't write them out. Instead, you would use up arrow notation to make the towers. Likewise, eventually you will get to a point where you will need a function to build towers of up arrows in the same way you needed the up arrows to build the exponentation towers. This function, also known as Conway Chains, already exists so it becomes level 4. Level 5 IS defined as the following: Every 5th function is called an Alpha Function, denoted with an A. A defines the instructions for building towers of Conway Notation. The first number in front of the A, called N, is the number repeated in the base of the Conway Tower an N amount of times. The number after the A, called X, defines the amount of towers of Conway Chains to build. For example, 5A1 means the following: 5->5->5->5->5 with one tower (so this is the correct expansion). Another example: 3A100 which is equal to 3->3->3...->3 with a 3A99 amount of chains in between. 3AA3 is 3A(3A(3A3))) using the rules of recursion. More advanced versions of A exist such as: 4AA(100)AA4 which contains 100 A's between the two 4s. 6th function is defined as beta. It does the same exact thing as A does to Conway Notation: 5B1 is 5AAAAA5, 3B100 is 3AA...AA3 with 3B99 A's in between. 3BB3 is 3B(3B(3B)), and 4BB(100)BB4 contains 100 A's between the two fours. The representation in NaN: 3A3 is (3{5,1}3) for the two 3s in the function, 5 for the level in which A is, and 1 is for having only one A. 3AA(100)AA3 is defined as (3{5,100}3). 3B3 is (3{6,1}3). Either way, in summary, for any particular greek letter, the corresponding NaN number will be the letter's place in the Greek alphabet plus 4. For any greek letter, it will tell you how to build towers of the previous level using the laws of recursion. JTOnstead20 (talk) 02:30, October 22, 2016 (UTC) : I could give you some insights on how to strengthen it while preserving its flavor and identity. But before i can do that, i must let you know the problem that makes NAN ill defined remains, it uses elements that do not form a part of a sequence. I'll be more specific, first, +, and */ ^ must not be separate elements, second, the recursion method you use isn't strong enough to bridge knuths arrows into conway arrows. Both are minor problems and easily corrected, i'll build an example for you on each of these two missing bridges, First lets say X is some binary operator/function and Y the first function above X in some ascending sequence You have 5XX3 = 5X(5X(5X5)) And 5Y3 = 5XXX3 Now lets assume X = +, what do you assume XX, XXX, XXXX is?, and then Y? 5XX3 = 5+(5+(5+5)) > 5*3. Ignore the use of > instead of = , the existance of some carry over in the final integers has little impact when comparing growth speeds of two functions. 5XXX3 = 5XX(5XX(5XX5)) > 5*5*5*5 > 5^3 5XXXX3 = 5XXX(5XXX(5XXX5)) > 5^5^5^5 > 5^^3 I hope it is clear now that given X=+ , aYb is on the power of a^...^b (ie. knuths). The second issue is a bit trickier to explain lets use my Y (at level of knuths numbered arrow) from just now and Z for what would be the next function after Y 10Y10 = 10->10 ->10 10YY10 = 10->10->10->2 10YYY10 = 10->10->10->3 10YYYYYYYYYY10 = 10->10->10->10 (note these werent exactly =) As you can probably Z now( :D ) , aZa is level a->a->a->a and not a->....->a .... There are plenty of ways to fix NAN, so its up to you. but the simplest would be to drop all the base functions except for -> and label that as your stage 1. You would need to use raw arrow chains and not towers of them, but "towers of chains of x" is not really much greater than "chains of x" (as a final note, i did neglect the fact that i am using 5B1 = 5A5, and not your 5AAAAA5, this difference would only make the explanation clunkier and the value shifts are negligible (with aZa using that mode being level a->a->a->a->2 ) ) (Chronolegends (talk) 05:49, October 22, 2016 (UTC)). I have compared a few things and realized that I can still resolve the issue: by making a comparison between NaN and FGH! You see, the growth rate of NaN on its first level is comparable to FGH levels. I will edit the first few levels of the NaN so that addition, multiplication, and exponentiation are all in level 1 of NaN just like in FGH. FGH follows like this: 1,2,3,4...w,w+1,w+2,w+3...w2,w3,w4... So, every time w upgrades, a new notation has been achieved. The 0,1 set is the first level of NaN as it includes addition, multiplication, and exponentiation. 2,3,4,5... is the next step as it includes all of up arrow notation. w+1, w+2, w+3 includes chained arrow notation, so that's level 3. Chained arrows are also included in w2, so w+1, w+2, w+3...w2 are all in level 3. Alpha notation would start at w^2 and end at w^w. Beta is at w^w...(w^w)+w, and so forth. I will revamp the notation system on my website and on the main page here by separating the older definition into informal and this newer definition into formal. JTOnstead20 (talk) 17:53, October 23, 2016 (UTC) nice, that fixes the ill-defined problem, NAN can treat the jump from 2 to 3 as an exception or special case and the rest of the levels follow a sequence. It looks a bit weird, but it works :D. Now for the fgh.., your values are a bit off track... but it'd be better if you figure those out on your own ;)' ' ps. can i add some NAN into my "numbers in CSBN" tables? (Chronolegends (talk) 18:58, October 23, 2016 (UTC)). Thank you. You can definetly add my notation to your table. But I need help: at least ten articles have the old, incorrect version of the NaN in the "Approximations" table. Can you help me find and correct those? Thank you. JTOnstead20 (talk) 19:07, October 23, 2016 (UTC) You got that wrong I'm sorry,but fφ(ω,0) is nowhere near the limit of FGH!Boboris02 (talk) 19:39, October 26, 2016 (UTC) I know that. fφ(ω,0) is the limit for NaN, but I meant the limit for the extension of NaN based on my ordinal notation called NEON. Please read the article of Ordinal Array Notation first before reading NEON, since Ordinal Array is the core portion of the extension. Thank you. JTOnstead20 (talk) 18:37, October 26, 2016 (UTC) OAN hits a snag at a fixed point starting with 3(0), and so is limited to n(0)=Γ_0. for all n>3 and btw rules 4 and 5 don't unstuck it. (Chronolegends (talk) 19:19, October 26, 2016 (UTC)). Yes,I did read the page,but to me it allways seems like the notation gets stuck at Γ₀.Boboris02 (talk) 19:40, October 26, 2016 (UTC) Oh, I see. I didn't define the numbers in between 3(0) and w(0). If that isn't the issue, then please say so. But I do have an explanation. The number in front of the parenthesis takes the limit of the function defined by the previous iteration of functions. It does not take an n amount of steps upon the number in parenthesis. For example: 1(0) is practically E0 because the 1 signifies taking the limit of 0 (omega) a single time. However, for the second time, we are not taking the second limit of omega itself (which would simplify to 2). Instead, we are taking the limit of the function 1(0) which can infinitize an ordinal a single time. Infinitizing the infiniting function yields the omega amount of ordinals, also known as phi(w,0). Likewise, when we move onto 3(0), we are not taking the infinitization of omega three times. Instead, we are taking the infinitization of the previous function, 2(0), which is to add omega in phi. Since the function is phi(x,0), then we infinitize phi itself to yield the Feferman ordinal. Now for the pressing part: 4(0). This is not equal to 4, the fourth infinitization of omega. Instead, it is equal to infinitizing the function 3(0), which was to make Feferman's ordinal. Due to the fact 3(0) takes phi notation itself to the limit, we are unable to use it to express the new ordinal created. However, extended phi notation tells us that the limit of phi(1,0) is phi(1,0,0). Thus, phi(1,0,0) can express the ordinal 4(0). The same thing goes with 5(0) (the Ackermann Ordinal) and so on to omega, where the next step takes hold. JTOnstead20 (talk) 02:50, October 27, 2016 (UTC) Omega has no limit, its an ordinal, not a function. Secondly, nesting phi(1,0,0) into phi(x,0) gives you phi(1,0,0) no matter how many times you nest (even if you do it infinite amount times, the result is still phi(1,0,0)=Γ₀ , this is what it means to be a fixed point) Thirdly, again, phi(1,0) is an ordinal, not a sequence or function, so phi(1,0) has no limit. "limit of phi (1,0)" has no meaning. (Chronolegends (talk) 21:34, October 27, 2016 (UTC)). I did not mean, in any cicumstance, to increase an ORDINAL to infinity. I meant to increase the function describing that particular ordinal to infinity. For example, w is defined as 1,2,3,4... I am not taking the infinitization of omega itself, rather, I am taking the infinitization of the set 1,2,3,4... As for phi(x,0): phi is a function, not an ordinal. The values produced by phi are ordinals but phi is in no way an ordinal in itself. By taking the infinity of phi(1,0), I am not taking the infinity of E0 (which it describes) but rather the phi function itself. Because infinitization means to increase somehting to omega, phi(1,0) becomes phi(w,0). As I said before, 3(0) means taking the infinity of the function ''2(0), not the ordinal described by 2(0). Since the function 2(0) is equal to phi(w,0), then take the infinity of the phi(w,0) function rather than the phi(w,0) ordinal. Lastly, for nesting phi(1,0,0) into phi(x,0): I never said the function 5(0) would nest phi(1,0,0) into phi(x,0). I do not see where phi(x,0) even came in to be a factor when concerning 5(0). Instead, I am taking the infinity of only phi(1,0,0) to get to phi(1,0,0,0). JTOnstead20 (talk) 02:41, October 28, 2016 (UTC) infinitization means to increase somehting to omega this definition is too vague to simply work for arbitrary inputs., but even if we take it "literally" the infinitization of the set 1,2,3,4... IS omega, not phi(1,0). The only thing that can "bridge" the gap between w and e0 is a sequence converging to w^^w, and if "infinitize x" is restricted to mean "do a sequence converging to x^^w", then 2(0) = e1. But you seem to think 2(0) is phi(w,0). The only thing that could brige that gap from e0 is the sequence converging to phi(1+w,0). but then if we restrict "infinitize x" to mean "do a sequence converging to phi(x+w,0) ," then 3(0)=phi(w*2,0). Before you respond, beware i am not talking about *your* infinitize. I'm trying to show you examples of what would be a well defined version, and why you can't simply slap some ordinals on a column, some strings on the right and expect to create a notation that "will extend to the limit FGH". which by the way, it can't TLDR; f_{phi(...)}(n) is too strong to be notationed into by the recursion methods you apply, "infinitize something" is not well defined, and NAN is still remains only well defined up to a{b,c}d '(Chronolegends (talk) 03:37, October 28, 2016 (UTC)).' I have decided to scrap everything and begin anew with OAN. But before I publish anything, even on my website, I want you to tell me if it will work or not. This new idea is based off something you said about tetration. Here it goes: Rule 1: OAN is a ordinal tetration notation. Every increase in a number can be represented by tetration of some sort. Rule 2: Omega is defined as 0 Rule 3: For all single numbers in the notation n, n+1 can be produced from n^^n. An example is this: 0 is omega and I want to increase it to 1. I have to tetrate a single time, so w^^w. This happens to be E0. Rule 4: The set n, n+1, n+2, n+3,... eventually reaches omega. When that occurs, add the second part of the notation to form k(n). In this case where the n set reaches omega, k will have advanced to its base level of 1. Rule 5: k can increase from its base level and advance by one through tetration. K's tetration is represented as k(n)^^k(n) = k+1(n). If k=0, then do not worry about including it in tetration. If k=1, then tetrate normally. Rule 6: The set k, k+1, k+2,... eventually reaches omega. When this occurs, add a second number x after a comma to get: n,x. The base level of n is 1. Rule 7: In order to increase x by one, the set k, k+1, k+2,... much reach omega in the expression k(n,x). Remember that k can increase through tetration. See step 5 for that. But now x is being included. Please review this notation and write a simple response saying either "yes" or "no" at the beginning, and a short explanation of why. Thank you. JTOnstead20 (talk) 13:49, October 28, 2016 (UTC) Nope but almost, youre just little bit off: hmm.. just because w^^w -> e0 and e0^^w -> e1 you can't simply infer "x^^x -> (x+1) " Instead of a lengthy boring explanation i'll just hand you some shortcuts. "x^^w depending on the value of x, x w e0>x>w (ie. x is some finite combination of omega and +*^) then x^^w -> e0 e_y+1 x >e_y (x is some finite combination of e_y and +*^) then x^^w -> e_y+1 x = some fixed point of e, such that e_x = x then x^^w -> e_x+1 (note, this also works for a combination of such a fixed point and +*^) And by the way, be careful when using tetration on ordinals, pay attention to the fact that i never use anything other than stuff^^w (ie. power tower of stuff that eventually rises to infinite height). Stuff like x^^(w+1) is not canonically defined and (x+1)^^y is not defined ordinals x with finite heights y. '(Chronolegends (talk) 16:21, October 28, 2016 (UTC)). ' I am not using an other number x. When I said n^^n, I meant put the ordinal described by n on both sides of the tetration. That means if n=E0 meand E0^^E0 not X^^E0. JTOnstead20 (talk) 19:27, October 28, 2016 (UTC) E0^^E0 has no standard / canonical definition or meaning, you could create or use an extension of ^^ before hand, but that is an even trickier mess than the one you are in right now. (While i was using stuff^^w, i explained to you exactly that it meant "power tower of stuff that eventually rises to infinite height" in the footnote., otherwise my shortcuts would also have been invalid) '(Chronolegends (talk) 19:47, October 28, 2016 (UTC)).' I have heard of E(E(E(E(E(E...)...) before. Isn't that equal to E0 to the power of itself E0 amount of times? Aside from that, say I do want to make a rule stating what "ordinal tetration" is. How would I define it so that E0^^E0 would equal E(E(E(E(E(E...)...)? JTOnstead20 (talk) 01:21, October 29, 2016 (UTC) There exists no work that would suggest E0^^E0 = E(E(...E...)...)? nor is it intuitively anywhere close to it., You have to understand, ordinals do some weird stuff even under normal arithmetic, for example are you aware of additive left cancellation ? w^7+w^8=w^8 < w^8+w^7 e0^^e0 under a custom ^^ extension that i built back in may points to e_e_0 but this is highly speculative and custom work and by no means should you take it as hint or advice, though i'll link to it just in case you might find educational or entertainment value on it.my ordinal tetration build '(Chronolegends (talk) 01:54, October 29, 2016 (UTC)).' Interesting, I may look into that further. In the meantime, I want you to tell me what my newest idea will do: Instead of having a tetration based system with w^^w as an example, I will have an up arrow notation equivalent with the example being w^^^...^^^w with a w amount of up arrows. This will change the notation rule from n+1= n^^n to n+1= n^^^...^^^n with n up arrows. Please tell me if this could work instead. JTOnstead20 (talk) 02:34, October 29, 2016 (UTC) nope, n^^^.....^^.n.. eventually reduces to some huge chain of n^^n so you can't avoid it with that technique'(Chronolegends (talk) 02:42, October 29, 2016 (UTC)).' I have a better idea then. I re-read the Epsilon article and found this: "the first fixed point of the function a->w^a." Then this is the definition is as follows: n+1 is equivalent to the first fixed point of the function a->n(a). Lets see how that works out. JTOnstead20 (talk) 03:11, October 29, 2016 (UTC) I removed your *tex because it looks like it bugged up this page, are you using some converter?, btw, all you are doing is shifting the problem of vague definition from the argument to the function. n(a) isn't defined before you begin talking about its fixed points, so you cant just say n+1 is the first fixed point of a->n(a) before you define what n(a) means. Chronolegends (talk) 04:47, October 29, 2016 (UTC) Okay. This is my last try at this. I hope it works. An ordinal can be defined not just by a function, but also as a set. For example, omega is defined by the set 1,2,3,4... Using this, I can make my rule: the set (w+1, w+2,... w2, w3,... w^2, w^3,... w^w, w^w^w, ...) defines E0. If n is an ordinal, then use successorship to create a set that can define n+1. A generalization of the set is (n, n+1, n+2,... n2, n3, ... n(2), n(3), ... n(n), n(n(n)),...) = n+1. So the formal rule is this: n+1 = the ordinal defined by the set (n, n+1, n+2,... n2, n3, ... n(2), n(3), ... n(n), n(n(n)),...). I am pretty confident that using the plus one trick will do the trick. So will it? JTOnstead20 (talk) 05:00, October 29, 2016 (UTC) Nope, only limit ordinals like have a fundamental sequence(set)...., successor ordinals don't. Please don't let my responses discourage your efforts, it is the opposite of what i would like. I know it can be frustrating or confusing, after all its a class of infinities we are talking about but i'm confident your persistence indicates you can learn You just need to research a bit more before trying to invent new ordinals. (and no, just rewording what you are doing is not proper research) Why not extend the greek function / NAN without them?, you were doing fine before the Z/OAN/NEON parts kicked in.'Chronolegends (talk) 05:27, October 29, 2016 (UTC).' Alrighty then. I give in. I will devise a new extension of NaN based on recursion at some point. I am not sure what to do with the OAN page, but I guess it is okay to delete it for now. I will be sure to let you know when the new version of NaN comes out so that I can be sure I am doing the right thing. Anyways, thank you for your support even though OAN did not work out. JTOnstead20 (talk) 06:08, October 29, 2016 (UTC) I will try to explain this notation idea to you, but if you have a better way of writing it, then please say so. I call it complex recursion. Let's say you have a function (a{x,y}b). Recursion will only take you to (a{x+1,y}b). However, say you have the ability to add another number (Z) that will compound the recursion! If z=1, then it means to compound one time which already happens. This means that, normally, z does not appear in the equation. Here are all the rules governing Z: 1. If Z=1, leave it out of NaN 2. If Z=2, then for any (a{X,Y,2}b), a new function (a{X,Y}b) will be created with X equal to the function with Z in it. An example is this: (10{5,5,2}10) becomes (10{X,5}10) with X equal to (10{5,5}10). 3. If Z=3, then (a{X,Y,2}b) will be repeated an (a{X,Y,2}b) amount of times 4. For any Z, it will equal (a{X,Y,Z-1}b) repeated a (a{X,Y,Z-1}b) times repeated a (a{X,Y,Z-1}b) amount of times. This continues a (a{X,Y,Z-1}b) amount of times, this entire operation (a{X,Y,Z-1}b) amount of times and so on until Z runs down to 0. Also define how fast this grows in FGH. Thank you. JTOnstead20 (talk) 02:06, October 30, 2016 (UTC)+ I'll only tell you this much: it works. Its not crazy strong, but i'll leave the work of finding out just how strong it is to you :D 'Chronolegends (talk) 03:05, October 30, 2016 (UTC)' I'm pretty sure my function only expands into the F_E(n) range. But at least it works. I will just build on this until I can reach a final conclusion. If you have any ideas, please feel free to share them. Thank you. JTOnstead20 (talk) 00:46, October 31, 2016 (UTC) .... it is nowhere close to epsilon level. Being "pretty sure" is worthless if you haven't got any solid grounds a.k.a. having hard checked enough values to be "pretty sure" in the first place. Plus, ive already warned you three times your approximation was off, NAN is nowhere near Epsilon level even with your proposed third bracket number extension. All letters use the same recursion style which i hereby call "tower recursion". You seem to have ignored the fact that conways only builds towers of knuths when doing chains of the kind x -> x -> x-> 2 , after this , conways surpasses the notion of knuthation towers. x->x->x->x is not feasible in towers for large X (see the picture in conway's tetratet for an example, it only gets worse when x=5 even at length 4 chain, and a length 5 becomes completely impossible to draw on a picture using knuth towers). As you can see, tower recursion is a weak method that does not bride you into huge numbers, which is a reason conway's chained arrow notation completely ditches tower recursion to begin with. Perhaps you will understand it better if i do your work for you and show you what actual FGH levels NAN reaches. they build upon towers of the last letter. Letting X be any letter, and Y the next letter above in the sequence, AY1 = AXAX....A with A Xs and AY2 = AXA...XA using AY1 X's nA1 feeds length of conways arrows so its fw^2(n) level nA2 feeds A1 into itself , fw^2(fw^2(n)) level nA100 is fw^2(....fw^2(n)...) level with 100 nestings nA(nA3) is fw^2+1(fw^2+1(3))(n) level nAA1 is fw^2+2(n) level nAA2 is fw^2+2(fw^2+2(n)) level nAAA1 is fw^2+3(n) level nAAA2 is fw^2+3(fw^2+3(n)) level nAAA100 is fw^2+3(fw^2+3(....fw^2+3(n) ....) ) level with 100 nestings nAAAA1 is fw^2+4(n) level nB1 is fw^2+w(n) level nB2 is fw^2+w(fw^2+w(n)) level nBB1 is fw^2+w+1(n) level nG1 is fw^2+w*2(n) level nGG1 is fw^2+w*2+1(n) level nD1 is fw^2+w*3(n) level nDDDDDDD1 is fw^2+w*3+6(n) level nD1 is fw^2+w*3(n) level nW1 is fw^2+w*23(n) level 100{100,100}100 is about fw^2+w*95+100(100) level (upper bound) 250{250,250}250 is about fw^2+w*245+250(250) level (upper bound) Caps at n{n,n}n reaching about f(w^2)*2(n) Proposed 3rd bracket number n{n,n,1}n beings at the end of two bracket number, f(w^2)*2(n) level n{n,n,2}n , replaces the second number with itself. f(w^2)*2(f(w^2)*2(n)) level n{n,n,3}n replaces the second number with itself, n{n,n,2}n times, f(w^2)*2(....f(w^2)*2(n)...), with f(w^2)*2(n) nestings = f(w^2)*2+1(f(w^2)*2(n)) , so under f(w^2)*2+1(f(w^2)*2+1(n)) n{n,n,4}n = same as above, but using n{n,n,3}n replacements , so around f(w^2)*2+1...(f(w^2)*2+1(n)...) aka. f(w^2)*2+2(f(w^2)*2+1(n)) or under f(w^2)*2+2(f(w^2)*2+2(n)) so the limit reached looks around n{n,n,n}n = f(w^2)*2+w(n) if we copy the extension such that n{n,n,n,X}n nests n{n,n,n}n into itself and so on, n{n....n}n is at the level f(w^2)*3(n) So as you can see NAN outpaced by peter hufords labeled arrows, 4 →_5 4 (around f(w^2)*4(4)) describes a number that can't be approximated with the current power of NAN 'Chronolegends (talk) 07:04, November 3, 2016 (UTC)''' You are missing one key part of NaN. Do you still build exponentiation towers on the fourth up arrow? Do you build Knuth Arrow notation towers on the a->a->a->3 of Chained arrow notation? Only the very first iteration of the notation builds towers of the previous notation. I have already matched NaN to FGH so if you want please check that out under "Formal Definition" in the NaN article. Thank you. JTOnstead20 (talk) 19:05, November 3, 2016 (UTC)